6. HCF and LCM of two numbers are 7 and 140 respectively. If the numbers are between 20 and 45, the sum of the numbers is:
a) 70 b) 63 c) 77 d) 56
Correct Answer: (b)
Let the numbers be 7x and 7y
where x and y are co-prime.
Now, LCM of 7x and 7y = 7xy
∴ 7xy = 140 ⇒ xy = 140/7 = 20
Now, the required values of x and y whose product is 50 and are coprime, will be 4 and 5.
∴ Numbers are 28 and 35 which lie between 20 and 45.
∴ Required sum = 28 + 35 = 63.
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7. The product of the LCM and the HCF of two numbers is 24. If the difference of the numbers is 2, then the greater of the number is
a) 3 b) 6 c) 4 d) 8
Correct Answer: (b)
Using Rule 1,
1st number × 2nd number = L.C.M. × H.C.F
Let the larger number be x.
Smaller number = x – 2
First number × Second number = HCF × LCM
→ x (x – 2) = 24
→ x2 – 2x – 24 = 0
→ x2 – 6x + 4x – 24 = 0
→ x (x – 6) + 4 (x – 6) = 0
→ (x – 6) (x + 4) = 0
→ x = 6 because x ≠ – 4
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8. The sum of the two numbers is 84 and their HCF is 12. The total number of such pairs of number is
a) 2 b) 4 c) 3 d) 5
Correct Answer: (c)
HCF = 12
∴ Numbers = 12x and 12y where x and y are prime to each other.
∴ 12x + 12y = 84
⇒ 12 (x + y) = 84
⇒ x + y = 84/12 = 7
∴ Possible pairs of numbers satisfying this condition
= (1,6), (2,5) and (3,4). Hence three pairs are of required numbers.
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9. L.C.M. of two numbers is 120 and their H.C.F. is 10. Which of the following can be the sum of those two numbers?
a) 140 b) 60 c) 80 d) 70
Correct Answer: (d)
Let the numbers be 10x and 10y
where x and y are prime to each other.
∴ LCM = 10 xy
→ 10xy = 120
→ xy = 12
Possible pairs = (3, 4) or (1, 12)
∴ Sum of the numbers = 30 + 40 = 70
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10. The sum of the two numbers is 384. H.C.F. of the numbers is 48. The difference in the numbers is
a) 100 b) 288 c) 192 d) 336
Correct Answer: (b)
Let the numbers be 48x and 48y
where x and y are co-primes.
48x + 48y = 384
→ 48 ( x + y) = 384
→ x + y = 384/48 = 8 …. (i)
Possible and acceptable pairs of x and y satisfying this condition are : (1, 7) and (3, 5).
∴ Numbers are : 48 × 1 = 48 and 48 × 7 = 336 and 48 × 3 = 144 and 48 × 5 = 240
∴ Required difference = 336 – 48 = 288
